OFFSET
1,2
COMMENTS
The minimum value of 5*a + 3*b is clearly 8 (a = b = 1), thus it is trivial that the Diophantine equation cannot be solved for any c < 8.
This sequence is finite. Proof: Given c > 7, we have three cases only to check (plus the particular cases c = 16 and c = 17 that we will see at the end), because 15 = 3 * 5. Thus, for every k > 5: 1) c := 3k --> so c = 5 * 3 + 5 * (k - 5), thus we can simply take (a = 3, b = k - 5); 2) c := 3k + 1 --> so c = 5 * 2 + 3 * (k - 3), thus (a = 2, b = k - 3); 3) c := 3k + 2 --> so c = 5 + 3 * (k - 1), thus (a = 1, b = k - 1). Now we can see that k > 5 is a sufficient but not necessary condition for the 2) and the 3), since in these cases it is sufficient that k > 3 which lead to a minimum value of 13 < 15. So we have proved that there is a solution for every c > 15.
Also, nonnegative numbers n such that the coefficient of x^n in the expansion of 1/((1-x^3)*(1-x^5))-1/(1-x^3)-1/(1-x^5) is zero. - Joerg Arndt, Aug 13 2015
LINKS
E. W. Weisstein, Diophantine Equation
EXAMPLE
For n = 1, 2, ..., 7 we have that 5*a + 3*b >= 8 > a(n).
CROSSREFS
KEYWORD
nonn,easy,fini,full
AUTHOR
Marco Ripà, Aug 07 2015
STATUS
approved